Solution 19
Physical Model:
When the container stops, the macroscopic kinetic energy of the gas, $K = \frac{1}{2}Mv^2$, transforms into internal energy. The problem considers a specific model for a di-atomic gas involving a characteristic temperature $T_0$ where rotational degrees of freedom (DOFs) activate.
- Range 1 ($T < T_0$): Rotational DOFs are frozen. The gas behaves like a monoatomic gas with $C_v = \frac{3}{2}R$.
- Transition ($T = T_0$): The rotational DOFs activate. This requires an absorption of energy equal to the rotational internal energy at that temperature ($U_{rot} = RT_0$) before the temperature can rise further. This acts like a “latent heat” phase.
- Range 3 ($T > T_0$): Rotational DOFs are fully active. The gas behaves as a standard di-atomic gas with $C_v = \frac{5}{2}R$.
Case I: Low Velocity ($T < T_0$)
The kinetic energy is insufficient to raise the temperature to $T_0$. The gas absorbs energy only in translational modes.
Energy Equation: $$ \frac{1}{2}Mv^2 = n C_{v1} \Delta T $$ $$ \frac{1}{2}Mv^2 = 1 \cdot \frac{3}{2}R (T – T_1) $$ $$ Mv^2 = 3R(T – T_1) $$ $$ T = T_1 + \frac{Mv^2}{3R} $$
Velocity Limit: This holds as long as $T \le T_0$. $$ T_1 + \frac{Mv^2}{3R} \le T_0 \implies v^2 \le \frac{3R(T_0 – T_1)}{M} $$ $$ v \le \sqrt{\frac{3R(T_0 – T_1)}{M}} $$
Case II: Intermediate Velocity ($T = T_0$)
In this range, the gas reaches $T_0$, and any excess kinetic energy is consumed to “unlock” the rotational degrees of freedom. The temperature remains constant at $T_0$ until the rotational energy requirement ($1 \cdot R \cdot T_0$) is fully met.
Final Temperature: $$ T = T_0 $$
Velocity Limit:
Lower limit: End of Case I, $v_1 = \sqrt{\frac{3R(T_0 – T_1)}{M}}$.
Upper limit: The total energy supplied equals the energy to reach $T_0$ (translational) plus the energy to activate rotation ($RT_0$).
$$ \frac{1}{2}Mv^2 = \frac{3}{2}R(T_0 – T_1) + RT_0 $$
$$ \frac{1}{2}Mv^2 = \frac{3}{2}RT_0 – \frac{3}{2}RT_1 + RT_0 $$
$$ \frac{1}{2}Mv^2 = \frac{5}{2}RT_0 – \frac{3}{2}RT_1 $$
$$ v^2 = \frac{R(5T_0 – 3T_1)}{M} \implies v_2 = \sqrt{\frac{R(5T_0 – 3T_1)}{M}} $$
Thus, for $\sqrt{\frac{3R(T_0 – T_1)}{M}} < v \le \sqrt{\frac{R(5T_0 - 3T_1)}{M}}$, the temperature is $T_0$.
Case III: High Velocity ($T > T_0$)
The kinetic energy is sufficient to heat the gas to $T_0$, fully activate rotational modes, and further heat the gas above $T_0$ using the higher heat capacity ($C_{v2} = \frac{5}{2}R$).
Energy Equation: $$ \frac{1}{2}Mv^2 = \Delta U_{total} $$ $$ \frac{1}{2}Mv^2 = \underbrace{\frac{3}{2}R(T_0 – T_1)}_{\text{Heat to } T_0} + \underbrace{RT_0}_{\text{Activate Rotation}} + \underbrace{\frac{5}{2}R(T – T_0)}_{\text{Heat above } T_0} $$
Simplifying the Right Hand Side (RHS): $$ \text{RHS} = \frac{3}{2}RT_0 – \frac{3}{2}RT_1 + RT_0 + \frac{5}{2}RT – \frac{5}{2}RT_0 $$ Grouping terms with $T_0$: $(\frac{3}{2} + 1 – \frac{5}{2})RT_0 = 0$. The $T_0$ terms cancel out. $$ \frac{1}{2}Mv^2 = \frac{5}{2}RT – \frac{3}{2}RT_1 $$
Solving for $T$: $$ \frac{5}{2}RT = \frac{3}{2}RT_1 + \frac{1}{2}Mv^2 $$ $$ 5RT = 3RT_1 + Mv^2 $$ $$ T = \frac{3T_1}{5} + \frac{Mv^2}{5R} $$
Condition: This applies when $v > \sqrt{\frac{R(5T_0 – 3T_1)}{M}}$.